Perpendicular Vector with given magnitude

[NEUhusky]

Homework Statement

Given a vector A = 8i − 5j, find the vectors in the xy plane that are perpendicular to A and have a magnitude of 14.

The Attempt at a Solution

I can find the vector perpendicular to A quite easily. I'm having trouble finding the perpendicular line when it has the given magnitude of 14. I have a feeling it's going to be an easy solution but something isn't clicking in my brain at the moment. Any help would be greatly appreciated. Thanks!

 

[tiny-tim]

Welcome to PF!

Hi NEUhusky! Welcome to PF!

The magnitude of ai + bj is √(a² + b²) …

show us what you get. ​

 

[NEUhusky]

Hi NEUhusky! Welcome to PF!

The magnitude of ai + bj is √(a² + b²) …

show us what you get. ​

Thanks! Glad I found this place!

Hm, well the magnitude of the given vector A is √(8^2+5^2) which is approximately 9.43.

Using the dot product I know that the perpendicular lines are either B = -5i-8j or 5i+8j because the dot product equals zero. Unfortunately the magnitude is still the same as A when it needs to be 14. Still kinda lost.

 

[tiny-tim]

ah!

no, 5i + 8j is a point on the perpendicular line …

multiply it by anything, and you'll get more points on the perpendicular line …

choose the correct multiples to get 14.

 

[NEUhusky]

ah!

no, 5i + 8j is a point on the perpendicular line …

multiply it by anything, and you'll get more points on the perpendicular line …

choose the correct multiples to get 14.

Ugh, still a bit confused. Multiples to get 14 would be either 2 & 7 or 1 & 14. I'm not sure how that applies. Doesn't the magnitude ( √(a2 + b2) ) have to equal 14?

 

[tiny-tim]

Hi NEUhusky!

(just got up :zzz: …)

Ugh, still a bit confused. Multiples to get 14 would be either 2 & 7 or 1 & 14. I'm not sure how that applies. Doesn't the magnitude ( √(a2 + b2) ) have to equal 14?

(try using the X² icon just above the Reply box )

Yes, but it doesn't have to be an integer!